Answer:
The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.
The null hypothesis is rejected.
There is significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units
Step-by-step explanation:
<u>Step:-(1)</u>
Given data the samples of material 1 gave an average (coded) wear of 85 units with a sample standard deviation of 4
Mean of the first sample x₁⁻ =85
standard deviation of the first sample S₁ = 4
Given data the samples of material 2 gave an average of 81 and a sample standard deviation of 5.
Mean of the first sample x₂⁻ =81
standard deviation of the first sample S₂ = 5
<u>Step :-2</u>
<u>Null hypothesis: H₀:</u> there is no significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units
<u>Alternative hypothesis :H₁: </u>there is significance difference between that the abrasive wear of material 1 exceeds that of material 2 by more than 2 units
Assume the populations to be approximately normal with equal variances.σ₁² =σ₂²
The test statistic

Given n₁=n₂=60.

On calculation, we get
Z =
z = 4.8389
The tabulated value Z =1.96 at 0.05 level of significance.
The calculated value Z= 4.8389> 1.96 at 0.05 level of significance.
The null hypothesis is rejected.
Conclusion:-
there is significance difference between that the abrasive wear of material 1 not exceeds that of material 2 by more than 2 units.
Answer:
10 cm
Step-by-step explanation:
Given:
No. of small spherical bulb = 1,000
radius (r) of smaller bulbs = 1 cm
Required:
radius of the bigger bulb
SOLUTION:
The following equation represents the relationship of the volume of the smaller and bigger bulb,

Where,
= volume of bigger bulb
= volume of smaller bulb
1,000 is the number of smaller bulbs melted to form the bigger bulb.
Volume of a sphere is given as, ⁴/3πr³
Therefore:
= ⁴/3*π*r³ = 4πr³/3
= ⁴/3*πr³ = ⁴/3*π*(1)³ = ⁴/3π*1 = 4π/3
Plug the above values into the equation below:





(12pie cancels 12 pie)
(taking the cube root of each side)
Radius of the bigger bulb = 10 cm
A = {1, 2, 5, 6, 8}
{1} U {2, 5, 6, 8}
{2} U {1, 5, 6, 8}
{5} U {1, 2, 6, 8}
{6} U {1, 2, 5, 8}
{8} U {1, 2, 5, 6}
{1, 2} U {5, 6, 8}
{1, 5} U {2, 6, 8}
{1, 6} U {2, 5, 8}
{1, 8} U {2, 5, 6}
{1, 2, 5} U {6, 8}
{1, 2, 6} U {5, 8}
{1, 2, 8} U {5, 6}
{1, 5, 6} U {2, 8}
{1, 5, 8} U {2, 6}
{1, 6, 8} U {2, 5}
The answer is 15 distinct pairs of disjoint non-empty subsets.